Question

The mid point of line joining the common points of the line $2x-3y+8=0$ and $y^2=8x,$ is

• A. $(3,2)$
• B. $(5,6)$
• C. $(4,-1)$
• D. $(2,-3)$

Answer 1

The correct option is B $(5,6)$

Let $(h,k)$ be the mid point.
Equation of chord with a mid point is $T=S_1$
$\Rightarrow yk-4(x+h)=k^2-8h\Rightarrow ky-4x=k^2-4h$
This line is identical with given line $2x-3y+8=0$

Comparing both the lines, we get
$\frac{k}{-3}=\frac{-4}{2}=\frac{k^2-4h}{-8}$
$\Rightarrow k=6$
and $16=k^2-4h\Rightarrow h=5$

Hence required mid point is $(5,6)$